Let $ E $ be the ellipsoid in $ \mathbb{R}^3 $ given by, $$ E = \left \{ (x,y,z) \mid \frac{x^2}{9} + \frac{y^2}{4} + z^2 = 1 \right \} $$ Find a smooth vector field $ H : \mathbb{R}^3 \rightarrow \mathbb{R}^3 $, such that $ H(v) $ is tangent to $ E $ for each $ v \in E $ and $ H $ vanishes at exactly $ 6 $ points of $ E $.
My idea : It suffices to find such vector field $ H $ for $ S^2 $, the unit sphere since there is an orthogonal transformation mapping $ E $ to $ S^2 $. Taking three mutually orthogonal planes through the origin would give us $ 6 $ good points on the sphere, for simplicity, take the coordinate planes $ x=0, y=0, z=0 $. Now, if $ n,s $ are the north and south poles, then planes passing through $ n,s $ and perpendicular to $ z=0 $ cut $ S^2 $ in great circles. For each such circle $ C $, consider $ C \backslash \{n,s\} $ and define $ F $, the field of unit vectors tangent to $ C $ on $ C \backslash \{n,s\} $. Thus $ F $ is defined on $ S^2 \backslash \{n,s\} $. Let $ H $ be the vector field obtained by multiplying the function $ (1-x^2)(1-y^2)(1-z^2) $ with $ F $ on $ S^2 \backslash \{n,s\} $ and $ 0 $ on $ n,s $. Does this work?
You are working too hard. The fact that the principal axes have different lengths is supposed to help you.
Hint: On the sphere, the unit normal vector field is always parallel to the vector field $x\partial_x + y\partial_y + z\partial_z$. But for a general surface this is not true.